(6d^(2)c)/(3c) what is the excluded value
1 answer:
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Answer:
a) P(X∩Y) = 0.2
b) = 0.16
c) P = 0.47
Step-by-step explanation:
Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.
So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67
Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:
P(X∩Y) = P(X) + P(Y) - P(X∪Y)
P(X∩Y) = 0.36 + 0.51 - 0.67
P(X∩Y) = 0.2
On the other hand, the probability that he must stop at the first signal but not at the second one can be calculated as:
= P(X) - P(X∩Y)
= 0.36 - 0.2 = 0.16
At the same way, the probability that he must stop at the second signal but not at the first one can be calculated as:
= P(Y) - P(X∩Y)
= 0.51 - 0.2 = 0.31
So, the probability that he must stop at exactly one signal is:
Answer:
A
Step-by-step explanation:
An excluded value is any value of x that makes the denominator of the rational expression zero as this would make the expression undefined.
For expression A
The denominator will be zero when x² - 4 = 0
x² - 4 is a difference of squares, thus
(x - 2)(x + 2) = 0
x - 2 = 0 ⇒ x = 2
x + 2 = 0 ⇒ x = - 2
The excluded values of x are x = ± 2 ⇒ A